LINK TO STANDARD MOLAR ENTROPIES:
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Standard-Molar-Entropies-984.html

LINK TO THERMODYNAMICS TABLE
http://www.genchem.net/thermo/thermotbl.html



ORANGE
a) delta G rxn = delta H rxn - T (delta S rxn)
2900J = 1900J - T (-3.36J/mol)
T= 297.6K or 24.45 degrees Celsuis
(you get Celsius by subracting 273.15 from the T in K)

b) Diamond thermodynamically should break down to graphite at room temperature and pressure but it does not because it is kinetically hindered. The tetrahedral arrangement of atoms in diamond have to be rearranged into hexagonal sheets of graphite and that is very unlikely. It requires a large activation energy (energy "hump" to be surmounted) which is simply not available at room temperature. So no, diamonds will not spontaneously convert to coal on your birthday. (from Yahoo! answers)


c) delta G rxn = delta H rxn - T (delta S rxn) T (in K) = 22 degrees C + 273.15

1900J - ( 295.15K) (-3.36 J/mol)

295.15K= 2891.704 J


YELLOW

You can see the answer in the file below. I couldn't get it to format properly on here and I didn't feel like sitting and fixing it for an hour. Hopefully it makes sense to you and hopefully I've done it correctly. - Deanna


for the calculation above, this is wat u put up:

"N2 +O2 = 2NO2
Srxn° = nS°(products) – nS°(reactants)
= [2( 239.9 J/molK)] – [( 191.5 J/molK) + 2( 205.0 J/molK)]
= [479.8 J/molK] – [191.5 J/molK + 410.0 J/molK]
= [479.8 J/molK] – [601.5 J/molK]
= -121.7 J/molK"

d) 5.0086*10^-19

uh... y is it 2( 205.0 J/molK) in the second step? There is only 1 mol of each reactant - N2 , O2 - jiahua